Bài giảng: Kỹ thuật siêu cao tần (CT389)Chương II: Đồ thị Smith(Smith Chart)

Luong Vinh Quoc Danh 9Basic Smith Chart techniques for loss-less transmission lines Introduction Cont’d  Given Zd  Find Γd Given Γd  Find Zd  Find d max and dminmaximum and minimum l

Luong Vinh Quoc Danh 1

Chương II : Đồ thị Smith

(Smith Chart)

Giảng viên: GVC.TS Lương Vinh Quốc Danh

Bộ môn Điện tử Viễn thông, Khoa Công Nghệ

E-mail: lvqdanh@ctu.edu.vn

Introduction:

 The Smith chart is one of the most useful graphical tools for high frequency

circuit applications.

 From a mathematical point of view, the Smith chart is a representation of all

possible complex impedances with respect to coordinates defined by the complex

reflection coefficient

 The domain of definition of the reflection coefficient for a lossless line is a circle

of unitary radius (vòng tròn đơn vị) in the complex plane (mặt phẳng phức) This is

also the domain of the Smith chart.

The goal of the Smith chart is to identify all possible

impedances on the domain of existence of the reflection coefficient.

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We start from the general definition of line impedance :

Introduction: (Cont’d)

) ( 1

) ( 1 )

x

x Z

x Z

)(1)(

x

x x

) ( ) (

Z x Z

Z x Z x

1)()(

x z x

Where  r = Re() and  i = Im()

Normalized line impedance: z = r + jx (2.8)

r = R/Z0 : normalized resistance

x = X/Z0 : normalized reactance

Introduction: (Cont’d)

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Introduction: (Cont’d)

Now we can rewrite (2.2) as follows:

i r

i r j

j jx

1

(2.10)

The real part r gives:

2 2

2 2

) 1 ( 1

i r

i r

r

i r

Introduction: (Cont’d)

The result for the real part indicates that on the complex plane with coordinates

(Re(Γ), Im(Γ)) all the possible impedances with a given normalized resistance r

are found on a circlewith

As the normalized resistance r varies from 0 to ∞ , we obtain a family of circles

completely contained inside the domain of the reflection coefficient | Γ | ≤ 1

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The imaginary partxgives:

2 2

) 1 ( 2

i r

i x

i r

Equation of a circle

(2.15)

The result for the imaginary part indicates that on the complex plane with

coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given normalized

reactance x are found on a circle with

Introduction: (Cont’d)

As the normalized reactancexvaries from -∞ to ∞ , we obtain a family of arcs

contained inside the domain of the reflection coefficient | Γ | ≤ 1

Introduction: (Cont’d)

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Basic Smith Chart techniques for loss-less transmission lines

Introduction (Cont’d)

 Given Z(d)  Find Γ(d)

Given Γ(d)  Find Z(d)

 Find d max and dmin(maximum and minimum locations for the voltage

standing wave pattern)

 Find the Voltage Standing Wave Ratio (VSWR)

 Given Z(d)  Find Y(d)

Given Y(d)  Find Z(d)

Phillip Hagar Smith (1905–1987): graduated from Tufts

College in 1928, invented the Smith Chart in 1939 while

he was working for the Bell Telephone Laboratories.

Smith Chart

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Scale in Wavelengths

Full Circle is One Half Wavelength Since Everything Repeats

Smith Chart (Cont’d)

Smith Chart (Cont’d)

(2.20)

where d = l – x

 = 0 (lossless)

(2.21)

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Smith Chart (Cont’d)

SWR : Standing Wave Ratio

RFL COEFF P : Reflection Coefficient Power || 2

RFL COEFF E or I : Reflection Coefficient Voltage ||

RTN LOSS [dB] : Return Loss S 11

Smith Chart : Admittances

j

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Since related impedance and admittance are on opposite sides of the

same Smith chart, the imaginary parts always have different sign.

Therefore, a positive (inductive) reactance corresponds to a negative

(inductive) susceptance, while a negative (capacitive) reactance

corresponds to a positive (capacitive) susceptance.

Analytically, the normalized impedance and admittance are related as

Smith Chart : Admittances (Cont’d)

Smith Chart : Nodes and Anti-nodes

Basic Applications

 Given Z(d) ⇒Find Γ(d)

1 Normalize the impedancez

2 Find the circle of constant normalized resistance r

3 Find the arc of constant normalized reactance x

4 The intersection of the two curves indicates the reflection

coefficient in the complex plane The chart provides directly the

magnitude and the phase angle ofΓ(d)

Example: Find Γ(d), given Z(d)=25+ j100 [Ω] with Z 0 = 50Ω

Basic Applications (Cont’d)

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Basic Applications (Cont’d)

• Given Γ(d) ⇒ Find Z(d)

1 Determine the complex point representing the given

reflection coefficient Γ(d)on the chart

2 Read the values of the normalized resistance r coefficient

point and of the normalized reactance x that correspond to the

reflection coefficient point

3 The normalized impedance is

and the actual impedance is

 Given Γ Rand Z R⇐⇒Find Γ(d) and Z(d)

1 Identify the load reflection coefficient ΓRand the normalized load

2 Draw the circle of constant reflection coefficient amplitude |Γ(d)|

= |Γ R |.

3 Starting from the point representing the load, travel on

the circle in the clockwise direction, by an angle

4 The new location on the chart corresponds to location d on the

transmission line Here, the values of Γ(d) and Z(d) can be read

from the chart as before

Basic Applications (Cont’d)

 Example: Given ZR = 25+ j 100 (Ω) with Z 0 =50 (Ω)

Find Z(d) and Γ(d) for d = 0.18λ

Basic Applications (Cont’d)

 Given Γ Rand Z R⇒Find d max and d min

1 Identify on the Smith chart the load reflection coefficient Γ Ror

the normalized load impedance Z R

2 Draw the circle of constant reflection coefficient amplitude |Γ(d)|

= |ΓR| The circle intersects the real axis of the reflection

coefficient at two points which identify dmax(whenΓ(d) = Real

positive) and dmin(when Γ(d) = Real negative)

3 A commercial Smith chart provides an outer graduation where

the distances normalized to the wavelength can be read directly

The angles, between the vector ΓRand the real axis, also provide

a way to compute dmaxand dmin

Basic Applications (Cont’d)

Example: Find dmaxand dmin for Z R =25+j100 (Ω); and Z R =

25−j100 (Ω) with Z 0 = 50 (Ω).

Basic Applications (Cont’d)

Basic Applications (Cont’d)

 Given Γ Rand Z R⇒Find the Voltage Standing Wave Ratio

(VSWR)

1 Identify the load reflection coefficient ΓRand the normalized

load impedance Z R on the Smith chart.

2 Draw the circle of constant reflection coefficient amplitude |Γ(d)|

= |ΓR|

3 Find the intersection of this circle with the real positive axis for

the reflection coefficient (corresponding to the transmission line

location dmax)

4 A circle of constant normalized resistance will also intersect this

point Read or interpolate the value of the normalized resistance

to determine the VSWR.

Basic Applications (Cont’d)

Basic Applications (Cont’d)

(Ω) with Z 0 = 50 Ω

 Given Z(d) ⇐⇒Find Y(d)

1 Identify the load reflection coefficient ΓRand the normalized

load impedance Z R on the Smith chart.

2 Draw the circle of constant reflection coefficient amplitude

|Γ(d)| =|ΓR|

3 The normalized admittance is located at a point on the circle of

constant |Γ| which is diametrically opposite to the normalized

Basic Applications (Cont’d)

Basic Applications (Cont’d)

 Find circuit impedance

Given R = 50 ; C 1 = 10 pF; C 2 = 12 pF; L = 22.5 nH;

 = 10 9 rad/s Find Z.

Choose Z 0 = 50 

Basic Applications (Cont’d)

Z

(point A) (point B)

When || = 0, there is no reflected waves, therefore, no reflected power Power generated

by signal source is totally received by load

 For lossy transmission line:

Power consumption at load:

Impedance matching (Cont’d)

2 2

: incident power at load

PL : reflected power at load

P S

 : incident power at source

PS : reflected power at source

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Impedance matching using lumped elements

ZL

Matching circuit

Example: ZL= 10 - j40 (); Z0= 50 ;  = 109rad/s

Impedance matching: -circuit

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Impedance matching: -circuit (cont’d)

 L 2 = R0/1.64 = 30.5 nH

 jx 1 = 1- zt‘ = - j1.55  C1= 1/1.5R0= 13 pF

Impedance matching: -circuit (cont’d)

Z L = 10 – j40 []

R 0 = 50 

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Impedance matching: forbidden regions

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Impedance matching: Single stub (cont’d)

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Impedance matching: Single stub (cont’d)

An RF Filter (1.1 GHz)

 Example: Z0= 50 ; ZL= (100 + j100) [] Stub #1: l 1,

short-circuited line, d = 0.4 Stub #2: l 2; d12= 3/8 Find l1, l 2.

Impedance matching: Double stub method

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Impedance matching: Double stub (cont’d)

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Impedance matching: Double stub (cont’d)

 The advantage of this technique is the position of stubs

(d 12 and x) are fixed The matching are done by

changing the length of stubs

 d12= /8, /4, or 3/8

 The disadvantage of this technique is not all impedances

can be matched (forbidden regions depend on values of

d 12 and x).

Impedance matching: Double stub (cont’d)

3.1GHz Microstrip Solid-State Amplifier The amplifier uses two double-stub

matching networks to achieve conjugate matching between the 50Ω

source/load and the transistor.

Impedance matching: Double stub (cont’d)